Earth's magnetic dipole homework problem

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SUMMARY

The forum discussion centers on solving a homework problem related to Earth's magnetic dipole. The user successfully calculated part a, yielding a magnetic field strength of 6.2 x 10^-5 T using the formula B=(μ0*μ)/(2πz^3). However, the user encountered an issue with part b, where they calculated the current (I) to be 6.3 x 10^8 Amperes, which was flagged as incorrect by their homework tool. The correct calculation for part b, using the relationship μ=AI, results in I=6.28 x 10^8 Amperes.

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vinamas
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Homework Statement
The earth's magnetic dipole is 8.0×10^22Am2
a)What is the magnetic field strength on the surface of the earth at the earth's north magnetic pole? You can assume that the current loop is deep inside the earth. Ans=6.25x10^-5 T


b)Astronauts discover an earth-size planet without a magnetic field. To create a magnetic field at the north pole with the same strength as earth’s, they propose running a current through a wire around the equator. What size current would be needed?
Relevant Equations
B=(mu0*mu)/(2piz^3)
mu=AI A=Area , I = current
So I'm mainly having trouble with part b as I have successfully completed part a. In part b I get an answer of 6.3 x 10 ^8 Amperes which my homework tool flags as incorrect. I'm clueless as to where to go from there or why my approach is wrong?
 
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Can you show your calculations for both parts please? Thanks.
 
Sure for part a) B=(mu0*mu)/(2piz^3) = (10^-7*2*8*10^22)/(6371 x 10^3)^3 = 6.2 x 10^-5 T
part b) mu=AI , 8x10^22=(pi*(6371x 10^3)^2)(I) solve for I=6.28 x 10^8
 

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